A few days ago, a former student of mine, David, came back to me about Box-Cox tests in linear models. It made me look more carefully at the test, and I do not understand what is computed, to be honest. Let us start with something simple, like a linear simple regression, i.e.

Let us introduced – as suggested in Box & Cox (1964) – the following family of (power) transformations

on the variable of interest. Then assume that

As mentioned in Chapter 14 of Davidson & MacKinnon (1993) – *in French* – the log-likelihood of this model (assuming that observations are independent, with distribution ) can be written

We can then use profile-likelihood techniques (see here) to derive the optimal transformation.

This can be done in R extremely simply,

> library(MASS) > boxcox(lm(dist~speed,data=cars),lambda=seq(0,1,by=.1))

we then get the following graph,

If we look at the code of the function, it is based on the QR decomposition of the matrix (since we assume that is a full-rank matrix). More precisely, where is a matrix, is a orthonornal matrix, and is a upper triangle matrix. It might be convenient to use this matrix since, for instance, . Thus, we do have an upper triangle system of equations.

> X=lm(dist~speed,data=cars)$qr

The code used to get the previous graph is (more or less) the following,

> g=function(x,lambda){ + y=NA + if(lambda!=0){y=(x^lambda-1)/lambda} + if(lambda==0){y=log(x)} + return(y)} > n=nrow(cars) > X=lm(dist~speed,data=cars)$qr > Y=cars$dist > logv=function(lambda){ + -n/2*log(sum(qr.resid(X, g(Y,lambda)/ + exp(mean(log(Y)))^(lambda-1))^2))} > L=seq(0,1,by=.05) > LV=Vectorize(logv)(L) > points(L,LV,pch=19,cex=.85,col="red")

As we can see (with those red dots) we can reproduce the R graph. But it might not be consistent with other techniques (and functions described above). For instance, we can plot the profile likelihood function,

> logv=function(lambda){ + s=summary(lm(g(dist,lambda)~speed, + data=cars))$sigma + e=lm(g(dist,lambda)~speed,data=cars)$residuals + -n/2*log(2 * pi)-n*log(s)-.5/s^2*(sum(e^2))+ + (lambda-1)*sum(log(Y)) + } > L=seq(0,1,by=.01) > LV=Vectorize(logv)(L) > plot(L,LV,type="l",ylab="") > (maxf=optimize(logv,0:1,maximum=TRUE)) $maximum [1] 0.430591 $objective [1] -197.6966 > abline(v=maxf$maximum,lty=2)

The good point is that the optimal value of is the same as the one we got before. The only problem is that the -axis has a different scale. And using profile likelihood techniques to derive a confidence interval will give us different results (with a larger confidence interval than the one given by the standard function),

> ic=maxf$objective-qchisq(.95,1) > #install.packages("rootSolve") > library(rootSolve) > f=function(x)(logv(x)-ic) > (lower=uniroot(f, c(0,maxf$maximum))$root) [1] 0.1383507 > (upper=uniroot(f, c(maxf$maximum,1))$root) [1] 0.780573 > segments(lower,ic,upper,ic,lwd=2,col="red")

Actually, it possible to rewrite the log-likelihood as

(let us just get rid of the constant), where

Here, it becomes

> logv=function(lambda){ + e=lm(g(dist,lambda)~speed,data=cars)$residuals + elY=(exp(mean(log(Y)))) + -n/2*log(sum((e/elY^lambda)^2)) + } > > L=seq(0,1,by=.01) > LV=Vectorize(logv)(L) > plot(L,LV,type="l",ylab="") > optimize(logv,0:1,maximum=TRUE) $maximum [1] 0.430591 $objective [1] -47.73436

with again the same optimal value for , and the same confidence interval, since the function is the same, up to some additive constant.

So we have been able to derive the optimal transformation according to Box-Cox transformation, but so far, the confidence interval is not the same (it might come from the fact that here we substituted an estimator to the unknown parameter .

Thank you for this post. Just a small correction. There is a small typo in confidence interval formula. In order to compute similar results with R, you may use:

ic=maxf$objective-0.5*qchisq(.95,1)

with this revision:

Lower: #[1] 0.2203471

Upper: #[1] 0.6696117

Means, half of Chi-Sq value will produce similar results.